Remark 3. Recall a curve is defined to be a connected, $1$-dimensional $C^1$ manifold. It’s a non-trivial fact that every connected $C^1$ curve can be oriented, and has a parametrization, and $\Gamma - \gamma(I)$ will be at most one point. (The proof is within reach; but requires work. It’s in the appendix of Milnor’s book, for instance.)

Remark 4. Sometimes it is useful to parametrize curves, which aren’t oriented. We say $\gamma$ is a parametrization of $\Gamma$ if it satisfies the first two conditions in Definition 2.

Remark 6. If $F = (F_1, \dots, F_d)$, where $F_i:\Gamma \to \R$ are functions, then one often writes the line integral in the differential form notation as \begin{equation}\tag{L}\label{e:lint} \int_\Gamma F \cdot d\ell = \int_\Gamma F_1 \, dx_1 + \cdots + F_d \, dx_d = \int_\Gamma \sum_{i = 1}^d F_i \, dx_i. \end{equation}

Remark 8. A curve may have multiple parametrizations. Proposition 7 shows that the line integral defined by \eqref{e:lint} is independent of the parametrization.

Remark 12. If the line joining through $a$ and $b$ passes through the origin, then some care has to be taken when doing the above computation.

Remark 13. We will see later that gravity is a conservative force, and that the above line integral only depends on the endpoints and not the actual path taken.

Remark 15. Any curve has exactly two orientations. If $\hat \tau$ is one orientation of $\Gamma$, then the other must be $- \hat \tau$. Reversing the orientation (i.e. replacing $\hat \tau$ with $-\hat \tau$) negates all line integrals, but leaves arch length integrals unchanged. As a result, when doing arc length integrals we won’t require the curve to be oriented.

Remark 19. Clearly $\gamma$ parametrizes $\Gamma$ by arc-length if and only if $\gamma$ is a parametrization with $\abs{\gamma’} \equiv 1$.

Remark 21. Let $\gamma$ parametrize $\Gamma$ by arc-length. Physically $\gamma(t)$ represents the position of a particle traveling along the curve $\Gamma$ with constant speed $1$. If $\Gamma$ is curved, the particle will experience a force when moving along the $\Gamma$. The more curved $\Gamma$ is, the stronger the force. This leads to the notion of curvature described here.

Remark 25. Our previous definition of manifold (without boundary) required it to be locally a graph of a $C^1$ function. That is for every $a \in M$, we required that for some small open set $U \ni a$, $M \cap U = \set{ P (x, g(x)) \st x \in V}$ where $V \subseteq \R^d$ is open, $g$ is $C^1$, and $P$ is a rearrangement of coordinates. If we set $\varphi(x) = P(x, g(x))$ (and restrict $V$ to a ball), then we get our new definition. From this proposition we know that for manifolds (without boundary), the new definition is equivalent to the old definition.

The change in the new definition is now we allow $B(x^*, r)$ to be an open ball in $\R^d_{+}$. If $x^*_1 > 0$, then making $r$ smaller if necessary, $B(x^*, r)$ looks like a standard ball in $\R^d$. If, however, $x^*_1 = 0$, then $B(x^*, r)$ looks fundamentally different. It is now only “half” a ball. When $d = 1$, $B(0, r) \subseteq \R^1_{+}$ is the half open interval $[0, r)$. When $d = 2$, $B(0, r) \subseteq \R^1_{+}$ is the half disk $\set{x \in \R^2 \st \abs{x} < r, x_1 \geq 0}$. Points $a \in M$ where we have $x^*_1 = 0$ for one $\varphi$, will be points for which $x^*_1 = 0$ for any $\varphi$ (this needs proof!) and these points are defined to be points on the boundary of $M$.

Remark 26. There are two notions of boundary, and both of them have the same name! The first notion of boundary is for subsets of metric spaces which we saw earlier. The metric space boundary of $M$ is just the closure of $M$! The manifold boundary of $M$, however is not always empty. If $M$ is a $d-1$ dimensional manifold, then the boundary is a $d-1$ dimensional manifold when it’s non-empty.

Remark 31. We allowed rotation of coordinates above because otherwise the square will not be a piecewise $C^1$ curve.

Remark 33. Some authors allow curves to “loop back over themselves”, and call curves which don’t “simple curves”. We define curves to be (connected) one dimensional manifolds – they can’t loop back over themselves. So our notion of curves is what is called a “simple curve” in many books.

Remark 34. For curves, the points in the boundary are called end points. A curve can have at most two endpoints (this is a theorem, which needs proof as mentioned in Remark 3.) If a curve has two endpoints, $a$, $b$, then one can parametrize it by a $C^1$ function $\gamma \colon [0, 1] \to \Gamma$ such that $\gamma(0) = a$ and $\gamma(1) = b$ (this again needs proof, as mentioned in Remark 3.) One can now orient $\Gamma$ by choosing $\hat \tau = \gamma’ / \abs{\gamma’}$. (In this case $a$, $b$ are called the starting and ending points of $\Gamma$ respectively.)

Remark 37 (Orienting closed curves). To do a line integral over a closed curve, one needs to orient them. For closed curves in the plane, one can orient them by traversing them clockwise (or counter clockwise). For closed curves in higher dimensions, the notions clockwise and counter-clockwise aren’t apriori defined, and need a frame of reference.

Remark 40. If $F$ is conservative, then and $\Gamma_1$, $\Gamma_2$ are any two curves starting from $a$ and ending at $b$, then \begin{equation} \int_{\Gamma_1} F \cdot d\ell = \int_{\Gamma_2} F \cdot d\ell \,. \end{equation} Physically, the work done by a conservative force only depends on the initial and final points, and not the path taken.

Remark 42. Clearly any potential force is conservative.

Remark 45. The fact that the answer to the previous problem is finite is why the escape velocity of the earth is finite. That is, if you launch a body straight up at the surface of the earth so that it has the same initial kinetic energy as the answer to the previous problem, then it won’t ever return to earth.